Answer to Question #242219 in Statistics and Probability for gabs

Question

(i) C(X +Y,Z)= C(X,Z)+C(Y,Z), where X and Y are two random vectors of the same dimension and C(X,Z)=C(Z,X).

(ii) If m=n, V(X+Y) =V(X) + V(Y)+ 2C(X,Y).

Answer 1

Let

$X=\left(X_1 ,....X_n \right)$

$Y=\left(Y_1 ,....Y_n \right)$

$Z=\left(Z_1 ,....Z_n \right)$

be any random vectors on the same probability space $P=(\varOmega, \varSigma, \mu)$

Then covatiance matrix C(X+Y,Z)=

$E\left( (X+Y-E(X+Y))(Z-EZ)^T \right)=\\ E\left( (X+Y-EX-EY))(Z-EZ) ^T \right)=\\ E\left( (X-EX)+(Y-EY))(Z-EZ)^T \right)=\\ E\left( (X-EX)(Z-EZ)^T + (Y-EY)(Z-EZ)^T \right)=\\ E\left( (X-EX)(Z-EZ)^T \right)+E\left( (Y-EY)(Z-EZ)^T \right)=\\ C(X,Z)+C(Y,Z)$

​First identity is proved.

During proving property of random matrices V,W : E(V+W)=EV+EW have being applied.

2) For proving second statement we need in additional condition C(X,Y)=C(Y,X)

$V(X+Y)=C(X+Y,X+Y)=]by\space 1]=C(X,X)+C(X,Y)+C(Y,X)+C(Y,Y)=\\ =V(X)+V(Y)+C(X,Y)+C(X,Y)^T=\\ [by\space asymption \space C(X,Y)=C(Y,X)]= \\ V(X)+V(Y)+2\cdot C(X,Y)$